A mirror diagonal diamond is Program 53’s V-shape plus a mirrored bottom half — digits climb 1..n on both diagonals, then descend back to 1.
Remember
Rule: top V (1..n), then bottom mirror (n-1..1)
1 1
2 2
3 3
4 4
5
4 4
3 3
2 2
1 1 ← rows = 5
Unlike Program 53 (top V only), this adds a second outer loop from rows - 1 down to 1 — total lines = 2n - 1.
Approach
How to Solve It
Reuse the same row logic twice: ascend i = 1..rows, then descend i = rows-1..1.
Method
Idea
Best for
Top + bottom
Same i === j / i === k row builder, two outer loops
Classic diamond demos
Compact
Same logic with fewer rows (e.g. 3)
Quick dry-runs
Pseudocode
Pseudocode
function printRow(i):
line = ""
for j from 1 to rows: // left
append i if i === j else " "
for k from rows-1 down to 1: // right
append i if i === k else " "
print line
for i from 1 to rows: // top
printRow(i)
for i from rows-1 down to 1: // bottom
printRow(i)
Cheat sheet
Goal
Pattern
Top half
for (let i = 1; i <= rows; i++)
Bottom half
for (let i = rows - 1; i >= 1; i--)
Left diagonal
line += (i === j ? i : " ");
Right diagonal
line += (i === k ? i : " "); with k from rows-1..1
End the row
console.log(line);
Total lines
2 * rows - 1
Printing Numbers vs Starting a New Line
API
Effect
Use for
line += …
Stays on the same row
Each digit or space
console.log(line)
Ends the current row
After both diagonal halves finish
Append without a newline, then end the row once.
Try it
Live Preview
Change the size and the diagonal diamond updates instantly.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 9 lines
1 1
2 2
3 3
4 4
5
4 4
3 3
2 2
1 1
Trace
Worked Walkthrough
Trace key lines when rows = 5 — top climb, peak, then bottom mirror.
Half
i
Role
Printed row
Top
1
First V row
1 1
Top
5
Peak (once)
5
Bottom
4
First mirror
4 4
Bottom
1
Last row
1 1
The bottom loop starts at rows - 1 so the peak line is never printed twice.
Code
JavaScript Programs
Three complete programs: fixed rows = 5, prompt input, and a compact rows = 3 demo. Use View Output for samples, or Try It Yourself to edit and run live.
Example 1 — Fixed rows = 5
Top loop prints the V; bottom loop mirrors it from rows - 1 down to 1.
JavaScript
const rows = 5;
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
for (let i = rows - 1; i >= 1; i--) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
1. Top half builds the V. Same as Program 53 — digit on i === j and i === k, spaces elsewhere.
2. Bottom half mirrors it. Loop i from rows - 1 down to 1 with identical inner logic.
3. Peak prints once. Starting the bottom at rows - 1 avoids a double middle row.
Example 2 — User Input Rows
Read rows with prompt and validate before drawing.
JavaScript
const rows = parseInt(prompt("Enter number of rows:"), 10);
if (!Number.isFinite(rows) || rows < 1) {
console.log("Please enter a positive integer.");
} else {
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
for (let i = rows - 1; i >= 1; i--) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
}
1. Prompt and validate. Use parseInt and require rows >= 1.
2. Same diamond core. Only the source of rows changes — top and bottom halves match Example 1.
3. Entering 4. Seven lines total (2×4 - 1), peak digit is 4.
Example 3 — Compact rows = 3
Same top + bottom logic with fewer rows for a quick visual check.
JavaScript
const rows = 3;
for (let i = 1; i <= rows; i++) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
for (let i = rows - 1; i >= 1; i--) {
let line = "";
for (let j = 1; j <= rows; j++) {
line += (i === j ? i : " ");
}
for (let k = rows - 1; k >= 1; k--) {
line += (i === k ? i : " ");
}
console.log(line);
}
1. Shrink the size. Only rows changes — still top climb then bottom mirror.
2. Easier to trace. Five lines total — useful while learning the two outer loops.
3. Same peak rule. The middle digit still appears only once.
Edge Cases & Pitfalls
Check these before calling the solution done.
Peak skip
Start the bottom loop at rows - 1
Starting at rows duplicates the middle tip line.
Center skip
Right inner loop still starts at rows - 1
Same as Program 53 — including k = rows doubles the center column.
rows = 1
Single-row case
Output is just 1 — the bottom loop never runs.
Bad input
Validate parseInt
Check Number.isFinite(rows) and require rows >= 1 before the loops.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / compact (Examples 1, 3)
O(n²)
O(n) per row string
User input (Example 2)
O(n²)
O(n) per row string
About 2n - 1 lines, each scanning about 2n - 1 positions — still quadratic in n.
Remember
Key Takeaways
Rule: Program 53’s V on top, then mirror it downward.
Peak once: start the bottom loop at rows - 1.
Break the row: call console.log(line) only after both diagonal halves finish.
Complexity:O(n²) — ~2n lines × ~2n chars.
One line: print the diagonal V upward, then reuse the same row logic downward from n - 1.
Frequently Asked Questions
The first loop prints the top V-half (i = 1..rows). The second prints the bottom half (rows-1 down to 1), mirroring the shape.
Starting at rows would print the middle row twice. rows-1 skips the peak already printed by the top half.
Program 53 prints only the top V-half. Program 54 adds a second outer loop to mirror the same row logic downward.
line += appends on the same row. console.log(line) ends the row after both diagonal halves finish.
The left loop uses i === j for the main diagonal. The right loop uses i === k for the mirrored diagonal, with spaces elsewhere.
O(n²) for n rows because the diamond has about 2n-1 lines and each line scans about 2n-1 positions.
Yes — write a function that builds one row for a given i, then call it from both outer loops.
One line prints a single 1 — the top loop runs once and the bottom loop does not run.
🤔
Did you know?
Program 53’s V-shape becomes a full diamond by adding a second outer loop from rows-1 down to 1. Total lines = 2n-1 with about 2n-1 characters each — O(n²) overall.
